JEE MAIN - Mathematics (2022 - 27th July Morning Shift - No. 9)
The area of the smaller region enclosed by the curves $$y^{2}=8 x+4$$ and $$x^{2}+y^{2}+4 \sqrt{3} x-4=0$$ is equal to
$$\frac{1}{3}(2-12 \sqrt{3}+8 \pi)$$
$$\frac{1}{3}(2-12 \sqrt{3}+6 \pi)$$
$$\frac{1}{3}(4-12 \sqrt{3}+8 \pi)$$
$$\frac{1}{3}(4-12 \sqrt{3}+6 \pi)$$
Explanation
_27th_July_Morning_Shift_en_9_1.png)
$$\cos \theta = {{2\sqrt 3 } \over 4} = {{\sqrt 3 } \over 2} \Rightarrow \theta = 30^\circ $$
Area of the required region
$$ = {2 \over 3}\left( {4 \times {1 \over 2}} \right) + {4^2} \times {\pi \over 6} - {1 \over 2} \times 4 \times 2\sqrt 3 $$
$$ = {4 \over 3} + {{8\pi } \over 3} - 4\sqrt 3 = {1 \over 3}\left\{ {4 - 12\sqrt 3 + 8\pi } \right\}$$
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